To find the rate of change of the area of the rectangle with respect to time, we need to differentiate the area function with respect to $t$.

Step 1: Write the formula for the area

The area $A$ of a rectangle is given by the formula: $A = \text{Length} \times \text{Height}$ Given:

• Length = $5t + 5$
• Height = $\sqrt{t}$

Thus, the area is: $A(t) = (5t + 5) \times \sqrt{t}$

Step 2: Differentiate the area with respect to time

Now, we differentiate $A(t)$ with respect to $t$. To do this, we'll use the product rule, which states: $\frac{d}{dt} [f(t) \times g(t)] = f'(t) \times g(t) + f(t) \times g'(t)$ Here:

• $f(t) = 5t + 5$
• $g(t) = \sqrt{t} = t^{1/2}$

First, we find the derivatives of $f(t)$ and $g(t)$:

• $f'(t) = 5$
• $g'(t) = \frac{1}{2} t^{-1/2}$

Now, applying the product rule: $\frac{dA}{dt} = f'(t) \cdot g(t) + f(t) \cdot g'(t)$ $\frac{dA}{dt} = 5 \cdot \sqrt{t} + (5t + 5) \cdot \frac{1}{2} t^{-1/2}$

Step 3: Simplify the expression

Now, we simplify the terms: $\frac{dA}{dt} = 5 \sqrt{t} + \frac{1}{2} (5t + 5) t^{-1/2}$ We can rewrite $\frac{1}{2} (5t + 5) t^{-1/2}$ as: $\frac{1}{2} (5t + 5) t^{-1/2} = \frac{1}{2} (5t^{1/2} + 5t^{-1/2})$

Thus, the rate of change of the area is: $\frac{dA}{dt} = 5 \sqrt{t} + \frac{1}{2} \left( 5t^{1/2} + 5t^{-1/2} \right)$

Step 4: Final expression

So, the rate of change of the area with respect to time is: $\frac{dA}{dt} = 5 \sqrt{t} + \frac{5}{2} t^{1/2} + \frac{5}{2} t^{-1/2}$ Which can be further simplified if needed, but this is the general form for the rate of change.

Would you like more details on any of these steps, or is there anything else you'd like to explore? Here's a list of related questions you might find interesting:

1. How do you apply the product rule when the functions involve powers of $t$?
2. What does the derivative of a function represent in terms of geometry or physics?
3. How can you interpret the rate of change of the area in practical terms?
4. What happens if you differentiate a function with respect to time when the function involves trigonometric functions?
5. How do you find the instantaneous rate of change of a function at a specific time?

Tip: When using the product rule, always identify the two functions being multiplied and differentiate them separately before combining them using the rule!